Jan 17 Line Integrals

Basic idea

Let's say you want to find the mass of a rope that's not of uniform density. You have to travel along the path and integrate by the adding the masses of each small slice of the rope.

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We can represent the integral along the path as
$$\int_C f(x,y) \, \mathrm{d}s = \int_a^b f(g(t), h(t))\sqrt{(g'(t))^2 + (h'(t))^2}\, \mathrm{d}t $$
The only restriction is it has to be smooth, meaning $f'(x,y)$ is never 0.

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Instead of writing in terms of $\mathrm{d}s$, we can write it as
$$\int_C f(x,y)\,\mathrm{d}x = \int_C f(x(t),y(t))x'(t)\,\mathrm{d}t$$
Note that if y'(t) = 0 in the original equation, it's the same thing, so this is independent of y. We can do the same thing with x.
$$\int_C f(x,y)\,\mathrm{d}y = \int_C f(x(t),y(t))y'(t)\,\mathrm{d}t$$
We can generalize and write this:
$$\int_C M(x,y)\, \mathrm{d}x + \int_C N(x,y)\, \mathrm{d}y$$
That's usually shortened to
$$\int_C M(x,y)\, \mathrm{d}x + N(x,y)\, \mathrm{d}y$$ In order to evaluate this, you must turn both d_'s into the same thing, be it $\mathrm{d}x$ or $\mathrm{d}t$.

Work

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A major application of line integrals deals with the work done by moving through a force field. So, imagine a submarine moving around the violent ocean. The current at each point is different, so it's nontrivial to caluclate the work done by moving from one place to another.
Now, the work is the sum of all of the $\Delta{}W$s. When the area you're looking at is so small, there is a constant force acting upon a point, and the direction is constant. Recall from section 14.3 that the work down by a constant force along a direction $V$ is $F \cdot V$. Using that, we can say
$$\Delta{}w_i = F(u_i, v_i, w_i) \cdot < \Delta{}x_i, \Delta{}y_i, \Delta{}z_i >$$
$$ = M(u_i, v_i, w_i)\Delta{}x_i + N(u_i, v_i, w_i)\Delta{}y_i + P(u_i, v_i, w_i)\Delta{}z_i$$
So, $$W = \int_C M(x,y,z)\, \mathrm{d}x + N(x,y,z)\, \mathrm{d}y + P(x,y,z)\, \mathrm{d}z$$
Finally, we can say that
$$W = \int_C F \cdot T\, \mathrm{d}s$$
$T$ is the unit tangent vector, and recall we're taking the dot product of many constant-direction works, so we need the unit tangent vector.
In addition, this is equal to $\int_C F \cdot dr$, and $dr = dxi + dyj + dzk$.